HARMONIC ANALYSIS ON SO(3) CHRISTIAN REMLING These notes are meant to give a glimpse into non-commutative har- monic analysis by looking at one example. I will follow Dym-McKean, Fourier Series and Integrals, Sect. 4.8 – 4.13, very closely. 1. The group SO(3) Since Fourier analysis on finite abelian groups worked so well, we now get (much) more ambitious and discuss an infinite non-abelian group. Our example is the group of proper rotations on R3, now denoted by SO(3) (“special orthogonal group” – “special” just means that the determinant is equal to 1). So 3×3 t SO(3) = {g ∈ R : g g = 1, det g = 1}. Such a rotation g can be described by three parameters. For instance, if you know the axis of rotation (specified by a direction or a point on S2 or two angles) and the angle of rotation (one parameter), g is determined uniquely. Alternatively, a matrix g ∈ R3×3 has 9 entries, but the requirement that gtg = 1 gives 6 conditions on these entries, and again 9 − 6 = 3. (The condition that det g = 1 singles out one half of the matrices satisfying gtg = 1; it does not reduce the dimension.) Summarizing in fancy language and adding some precision, we have: Theorem 1.1. SO(3) is a (compact) 3-dimensional manifold (what- ever that means). We can’t use characters to analyze functions on G = SO(3). This does not come as a surprise because G is not commutative and a char- acter χ can’t distinguish between gh and hg: χ(gh) = χ(g)χ(h) = χ(hg) More to the point, it can be shown that the only character χ on SO(3) is the trivial character χ(g) ≡ 1. To analyze functions on G, we break G into smaller pieces. Let K be the subgroup of rotations about the z axis. Equivalently, K is the set of rotations that fix the north pole n = (0, 0, 1)t. An explicit description 1 2 CHRISTIAN REMLING of K is given by cos ϕ − sin ϕ 0 K = {k(ϕ) : 0 ≤ ϕ < 2π}, k(ϕ) = sin ϕ cos ϕ 0 . 0 0 1 K is a subgroup of G, and in fact K ∼= S1. Indeed, the map k(ϕ) 7→ eiϕ is an isomorphism from K onto S1. Exercise 1.1. Check this. We now introduce the cosets of K gK = {gk : k ∈ K} and the set of cosets G/K = {gK : g ∈ G}. (Warning for readers with some knowledge of group theory: K is not a normal subgroup and G/K is not a group.) Exercise 1.2. Prove that two cosets g1K, g2K are either equal or dis- joint. Given h ∈ G and a coset gK, the group element h acts on the coset gK in a natural way and produces the new coset hgK. The next theorem shows that the coset space G/K can be naturally identified with S2. Moreover, if looked at on S2, the above action becomes the map x 7→ hx (x ∈ S2, h ∈ SO(3)). Theorem 1.2. There exists a bijective map j : G/K → S2 so that j(hgK) = hj(gK) for all g, h ∈ G. Proof. Let n = (0, 0, 1)t be the north pole. We would like to define j(gK) = gn but before we can do this, we must check that the right- hand side is independent of the choice of the representative g. In other words, if g1K = g2K, then we must also have that g1n = g2n. Now if g1K = g2K, then g2 = g1k for some k ∈ K and since k fixes the north pole, g2n = g1kn = g1n, as desired. It is clear that j satisfies j(hgK) = hj(gK). Moreover, j maps G/K onto S2 because for every x ∈ S2, there exists a rotation g so that gn = x. It remains to show that j is injective. If g1n = g2n, then the −1 rotation g2 g1 fixes n and thus must be in K. But then g1K = g2K, so g1, g2 actually represent the same coset. We have already seen that we can let group elements act on cosets gK. We will now be especially interested in the double coset space K/G/K = {KgK : g ∈ G}, HARMONIC ANALYSIS ON SO(3) 3 where, as expected, KgK = {k1gk2 : k1, k2 ∈ G}. Things become very transparent if we use the identification G/K ∼= S2 from above. Then gK corresponds to a point x on S2, and k ∈ K acts on this by just doing the rotation kx. Now K is precisely the set of rotations about the z axis, so KgK ∼= Kx is a circle of constant latitude on the sphere. In particular, we can parametrize the elements of K/G/K by using this latitude θ. In other words, θ is the angle a vector pointing towards the circle makes with the z axis, and 0 ≤ θ ≤ π. 2. Integration on G We can’t make any serious progress without being able to integrate functions defined on G. There is heavy machinery that addresses this issue in a very general setting, but we don’t need any of this here. We just recall from the previous section that we can naturally identify G ∼= G/K × K and also G/K ∼= S2, K ∼= S1, and we do know how to integrate on S1 and S2, respectively. This then automatically gives us an integral on G. To carry out this program, associate with a (sufficiently nice) func- tion f : G → C its average f0 over gK: Z f0(g) = f(gk) dk K More precisely, we actually do the integral 1 Z 2π f(gk(ϕ)) dϕ, 2π 0 making use of the existing integration theory on S1 ∼= [0, 2π). However, at least for theoretical use of the integral, it’s usually better to be less explicit in the notation. The point is that f0 only depends on the coset gK of g, not on g itself. In a sense, this is clear because f0 was defined as the average over that coset. The formal proof depends on the (left and right) invariance of the integral on K: For every continuous (say) function f : K → C and k0 ∈ K, Z Z Z (2.1) f(k) dk = f(k0k) dk = f(kk0) dk. K K K Exercise 2.1. Prove (2.1). (The proof consists of unwrapping the defi- nitions.) 4 CHRISTIAN REMLING Now (2.1) indeed shows that for arbitrary k0 ∈ K, Z Z 0 0 f0(gk ) = f(gk k) dk = f(gk) dk = f0(g). K K This says that f0 is constant on every coset gK. In particular, making use of the identification G/K ∼= S2, we can define Z 1 Z (2.2) f(g) dg = dσ(x)f0(x). G 4π S2 Again, this is actually short-hand for the more precise formula Z Z 1 −1 f(g) dg = dσ(x)f0(j (x)), G 4π S2 where j−1 is the inverse of the identification map j : G/K → S2 from Theorem 1.2. Even this is not totally accurate, we would actually need the function fe0 : G/K → C induced by f0 : G → C in the integral. Of course, (2.2) is the version we’ll work with. The factor 1/4π makes sure that the integral is normalized in the R sense that G dg = 1. It is also left-invariant, that is, Z Z (2.3) f(hg) dg = f(g) dg. G G In fact, dg is the only measure on SO(3) with these properties. It is called the Haar measure. Exercise 2.2. Prove (2.3). Again, you will need to unwrap the defini- tions. The Haar measure on SO(3) has additional nice properties: Theorem 2.1. Let f : G → C a continuous (say) function and h ∈ G. Then Z Z Z Z f(g) dg = f(g−1) dg = f(gh) dg = f(hg) dg. G G G G Proof. Given f, define a new function f −1 by f −1(g) = f(g−1). Left- invariance of dg (see (2.3)) then shows that Z Z Z Z f(g) dg = f(h−1g) dg = dh dg f(h−1g) G G G G Z Z Z Z = dg dh f −1(g−1h) = dg dh f −1(h) G G G G Z Z = f −1(h) dh = f(g−1) dg. G G HARMONIC ANALYSIS ON SO(3) 5 Given this and left-invariance, the right-invariance now follows from the calculation Z Z Z f(gh) dg = f −1(h−1g−1) dg = f −1(h−1g) dg G G G Z Z Z = f −1(g) dg = f(g−1) dg = f(g) dg. G G G 3. Convolutions Recall that if X = S1 or X = Rd, then the Fourier transform is a linear map on the functions on X. Moreover, it also respects the convolution product of functions: (f ∗ g)b= fbgb . We will now look for similar maps on functions on G = SO(3). To do this, we must first define a convolution for functions f : G → C. The obvious try is Z −1 (f1 ∗ f2)(g) = f1(gh )f2(h) dh G (as usual, if in doubt, assume that f1, f2 are nice smooth functions; from a structural point of view, however, it would actually be best to work with the class L1(G) of merely integrable functions here). Exercise 3.1. Prove that convolution is associative. Unfortunately, convolution is not commutative on SO(3).